This question is succeeding a question raised in this post. Let $$f:X\dashrightarrow Y$$ be a rational map between smooth projective varieties over $\mathbb C$ with smooth fundamental locus $B$ and $\text{codim}_BX=2$. It is mentioned by Sasha that by considering the closure of graph $\overline{\Gamma(f)}$ inside the product $X\times Y$, the first projection $\pi_1:\overline{\Gamma(f)}\to X$ is a projective birational morphism, therefore the blowup along some ideal sheaf $\mathcal{I}$ supported on $B$ (Hartshorne Cha. II, 7.17) and the second projection $\pi_2:\overline{\Gamma(f)}\to Y$ becomes regular.
This is a way of extending a rational map to a morphism. Another way is to blowup smooth centers successively according to Hironaka's theorem. What I want to ask is something "converse", namely, I believe the following statement is true:
Claim: Let $\sigma:\text{Bl}_BX\to X$ be the blowup along the (reduced) subvariety $B$, and assume that the induced rational map $\tilde{f}$ on the blowup $$\tilde{f}:\text{Bl}_BX\to Y$$ is regular. Then $\text{Bl}_BX\cong \overline{\Gamma(f)}$.
In other words, what I want to prove is that if the rational map extends to the initial blowup $\text{Bl}_BX$, then the graph closure is isomorphic to blowup of the ideal sheaf $\mathcal{I}_B$ of $B$. What I have so far is that by the universality property of the blowup (Hartshorne Cha. II, 7.14), there is a unique morphism $g$ such that $\pi_1$ factors through $$\overline{\Gamma(f)}\xrightarrow{g} \text{Bl}_BX\xrightarrow{\sigma} X$$
However, to prove my claim, I also need to dominate $\overline{\Gamma(f)}$ by $\text{Bl}_BX$, somehow using the assumption that $\tilde{f}$ is a morphism, but I don't know how to go further.
Any suggestions and comments are appreciated!
Note that $\mathrm{Bl}_B(X)$ comes with the blowup morphism to $X$ and with the morphism $\bar{f}$ to $Y$. Together, they define a morphism $$ \mathrm{Bl}_B(X) \to X \times Y. $$ Its image is an irreducible and reduced subscheme of $X \times Y$ which contains the graph of $f\vert_{X \setminus B}$ as a dense open subset, hence coincides with $\overline{\Gamma(f)}$. This gives the require morphism $\mathrm{Bl}_B(X) \to \overline{\Gamma(f)}$.